FIGURE P12.63 shows a 15 kg cylinder held at rest on a 20° slope. b. What is the magnitude of the static friction force?

Video transcript

Little fellow physicist today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A cylindrical object with a mass 25 kg is positioned on a sloping surface that forms a 32 degree angle with the horizontal. The cylindrical object remains at rest on the 32 degree incline, determine the magnitude of static friction that prevents the cylindrical object from rolling down the inclined surface. OK. So first off, let us look at a diagram to help us better visualize this problem. So I've gone ahead and drawn over here on the right down below. From the problem. A diagram here it has a circle here and represent the cylindrical object. And it shows that it's being showing the static friction which is the curly f that prevents the cylindrical object from rolling down the incline surface. We also have some tension here because it's something's preventing between static friction and some like tension T is holding, you know the object from rolling down the incline and also has the cylindrical surface that's positioned that forms a 30 degree angle with the horizontal, which is shown. So here's the 32 degree angle, which also means when we draw like a, when we make like a little triangle here, and then we use some trig and all that with the 32 we get that FG equals cosine data considering this part of the triangle and that we get FG multiplied by sine theta for this length going to the left. And then considering, you know up is in the Y axis direction and then is the vertical and then horizontal is in the X axis direction. OK. So now let us note that since the cylindrical object stands still, it has translational and equal E Librium conditions or qualities. OK. So we need to recall that translational equilibrium states that F net the net force is equal to zero. And that rotational equilibrium states that the sum of torque equals zero. So if we apply the condition of translational equilibrium to the axis along the inclined plane, we can write that the sum of F subscript X FX is equal to zero. And if we refer to our diagram, we can take it even further to say that the sum of FX is equal to capital TT plus FS which is gonna be our static friction. OK? Minus M multiplied by G multiplied by sine data. OK. So we can go on and right that T plus the static friction minus MG where M is the mass and G is gravity multiplied by sine theta equals zero. So now we need to rearrange this equation to solve for the static friction. So fs the static fraction is equal to mass multiplied by gravity multiplied by sin theta minus T. And let's call this equation one. OK. So our end goal is to find the magnitude, but we must first determine the value for teeth in order to find the magnitude. So in order to find the unknown value for T, let us rearrange the condition of rotational equilibrium about the contact point on the cylindrical object with an inclined plane. Once we apply this condition, it will simplify our solving method. Since we do not need to consider the frictional force when we apply this condition. So we can write that the sum of torque equals zero. And that is the same as saying two R multiplied by T minus R multiplied by the mass M multiplied by G, the gravity multiplied by sine of degrees is equal to zero. So let's solve for tea, it rearranged. And so for T so T is equal to M multiplied by G multiplied by sine of 32 degrees divided by two. Now we can solve our T by plugging in all of our known variables. So let's do that. So T is equal to 25 kg. This M is equal to 25 kg multiplied by gravity which is 9.81 meters per second squared, multiplied by sign of 32 degrees all divided by two. So when we plug it into a calculator, we should get 64. newtons. OK. So now we can plug in our value for T back into equation one. So FS is equal to kg multiplied by sine of 32 degrees multiplied by gravity which was 9.81 m per second squared minus N or minus. I said N but I was meant to say minus T, which is an N Newton units which was the 64.92 newtons is T. So we could plug in T. So it's 25 kg multiplied by sin of 32 degrees multiplied by gravity. 9.81 m per second squared minus 64.92 newtons. So when we plug that into a calculator, they should get that the static friction is equal to 65. newtons. All right, we did it so we can just easily round up to 65 newtons to make it easy. So let's look at our multiple choice answers to see which answer is correct. The correct answer is the letter A 65 newtons. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.